Question

A particle moving in a circle of radius $R$ with a uniform speed takes a time $T$ to complete one revolution. If this particle were projected with the same speed at an angle $\theta$ to the horizontal, the maximum height attained by it is $4R$. The angle of projection, $\theta$ is then given by :

• A. $\theta ={\cos }^{-1}{\left(\frac{{\pi }^{2}R}{g{T}^2}\right)}^{1/2}$
• B. $\theta ={\sin }^{-1}{\left(\frac{{\pi }^{2}R}{g{T}^2}\right)}^{1/2}$
• C. $\theta ={\sin }^{-1}{\left(\frac{2g{T}^2}{{\pi }^{2}R}\right)}^{1/2}$
• D. $\theta ={\cos }^{-1}{\left(\frac{g{T}^2}{{\pi }^{2}R}\right)}^{1/2}$

Answer 1

The correct option is C $\theta ={\sin }^{-1}{\left(\frac{2g{T}^2}{{\pi }^{2}R}\right)}^{1/2}$

To complete a circular path of radius $R$, the time period is $T$.

So, speed of the particle, $v=\frac{2\pi R}{T}...\left(1\right)$

Now, the particle is projected with the same speed at angle $\theta$ to horizontal.

So, maximum height,

$H=\frac{v^2{\sin }^2\theta }{2g}$

Given, $H=4R$

$\Rightarrow \frac{v^2{\sin }^2\theta }{2g}=4R$

$\Rightarrow {\sin }^2\theta =\frac{8gR}{v^2}$

$\Rightarrow {\sin }^2\theta =\frac{8gR{T}^2}{4{\pi }^2{R}^2}=\frac{2g{T}^2}{{\pi }^{2}R}\left[\text{From equation}\right(1\left)\right]$

$\Rightarrow \theta ={\sin }^{-1}{\left(\frac{2g{T}^2}{{\pi }^{2}R}\right)}^{1/2}$

Hence, option $\left(A\right)$ is the correct answer.